The molar solubility of \( \text{Zn(OH)}_2 \) in \( 0.1 \, \text{M} \) NaOH sodium is \( x \times 10^{-18} \, \text{M} \). The value of \( x \) is:(Given: The solubility product of \( \text{Zn(OH)}_2 \) is \( 2 \times 10^{20} \) )